Electric field outside a long wire carrying charge q is proportional to :

\frac{1}{r}

\frac{1}{r^{2}}

\frac{1}{r^{\frac{1}{3 / 5}}}

\frac{1}{r^{\frac{1}{3 / 2}}}

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Solution

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Using Gauss Law across the cylindrical Gaussian surface,
$\int E . d S = \frac{q_{e n c}}{ϵ_{0}}$
$∴ E \times 2 π r \times l = \frac{λ \times l}{ϵ_{0}}$
$∴ E = \frac{λ}{2 π r ϵ_{0}}$
Here $λ$ is the linear charge density on the wire.

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Guides

Electric field outside a long wire carrying charge q is proportional to :

Using Gauss Law across the cylindrical Gaussian surface,∫E.dS=qencϵ0∴E×2πr×l=λ×lϵ0∴E=λ2πrϵ0Here λ is the linear charge density on the wire.

Using Gauss Law across the cylindrical Gaussian surface,
$\int E . d S = \frac{q_{e n c}}{ϵ_{0}}$
$∴ E \times 2 π r \times l = \frac{λ \times l}{ϵ_{0}}$
$∴ E = \frac{λ}{2 π r ϵ_{0}}$
Here $λ$ is the linear charge density on the wire.